Solution:

$(i) \dfrac{e^{-2s}}{s^2+8s+25}$

$L^{-1}[e^{-as}\phi(s)]=f(t-a)H(t-a) \\ where \ L^{-1}[\phi(s)]=f(t)$

$Let \ \phi(s) \ = \ \dfrac{1}{s^2+8s+25} \\[2ex] f(t)=L^{-1}[\phi(s)]=L^{-1} \dfrac{1}{s^2+8s+25} \\[2ex]$

$=L^{-1}\dfrac{1}{s^2+8s+16-16+25} \\[2ex] = L^{-1}\dfrac{1}{{\left(s+4\right)}^2+9} \\[2ex] =e^{-4t}L^{-1}\dfrac{1}{s^2+3^2} \\[2ex] =e^{-4t}\ \dfrac{1}{3}\ sin3t$

$f(t-2) = e^{-4\left(t-2\right)}\ \dfrac{1}{3}\ sin3(t-2)$

$L^{-1}\dfrac{e^{-2s}}{s^2+8s+25}= e^{-4\left(t-2\right)}\ \dfrac{1}{3}\ sin3(t-2).H(t-2)$

$(ii) \ \dfrac{e^{-3s}}{{\left(s+4\right)}^3}$

$L^{-1}[e^{-as}\phi(s)]=f(t-a)H(t-a) \\ where \ L^{-1}[\phi(s)]=f(t)$

$Let \ \phi(s)= \dfrac{1}{{\left(s+4\right)}^3}$

$f(t) = L^{-1}[\phi(s)] = L^{-1}\dfrac{1}{{\left(s+4\right)}^3} \\[2ex] =e^{-4t}\ L^{-1}\dfrac{1}{{\left(s\right)}^3} \\[2ex] =e^{-4t}\ \dfrac{t^2}{2} \\[2ex] =f(t-3) = e^{-4\left(t-3\right)}\ \dfrac{{\left(t-3\right)}^2}{2}$

$L^{-1}\dfrac{e^{-3s}}{{\left(s+4\right)}^3}= e^{-4\left(t-3\right)}\ \dfrac{{\left(t-3\right)}^2}{2}H(t-3)$